Final answer to the problem
$x>\frac{\sqrt{321}+70}{19}$
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1
Use the complete the square method to factor the trinomial of the form $ax^2+bx+c$. Take common factor $a$ ($19$) to all terms
$19\left(x^2-\frac{140}{19}x+\frac{241}{19}\right)>0$
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$19\left(x^2-\frac{140}{19}x+\frac{241}{19}\right)>0$
Learn how to solve problems step by step online. Solve the inequality 19x^2-140x+241>0. Use the complete the square method to factor the trinomial of the form ax^2+bx+c. Take common factor a (19) to all terms. Add and subtract \displaystyle\left(\frac{b}{2a}\right)^2. Factor the perfect square trinomial x^2+-\frac{140}{19}x+\frac{4900}{361}. Multiply the fraction and term in - \frac{70}{19}.
Final answer to the problem
$x>\frac{\sqrt{321}+70}{19}$
